2 Answers

(A) is false. Consider just two vertices connected to each other. So, we have one SCC. The new graph won't have any edges and so $2$ SCC.

(B) is true. In a directed graph an SCC will have a path from each vertex to every other vertex. So, changing the direction of all the edges, won't change the SCC.

(D) is false. Consider any graph with isolated vertices- we loose those components.

(C) is a bit tricky. Any edge is a path of length $1$. So, the new graph will have all the edges from old one. Also, we are adding new edges $(u,v)$. So, does this modify any SCC? No, because we add an edge $(u,v)$, only if there is already a path of length $<= 2$ from $u$ to $v$- so we do not create a new path. So, both (B) and (C) must answer, though GATE key says only B.

A. Can be eliminated as if the component is strongly connected in E then it will not be strongly connected in E's complement.

B.This is reversing the directions of the edges. When a triangle is considered for instance, where the directions are such that it will form a cycle, on reversing these directions the triangle will no more be strongly connected as there will be a vertex from which we cant reach out to any other vertex.

C.This is mostly the answer as according to the condition the number of edges in E3 will be those present in E and also additional edges which connect all U to all V such that UV<=2. Hence the connected component will stay connected.

D.V4 has only single vertices ,i.e those which were isolated in G and no edges will be present between them.

doubt: (D) also seems correct.. V4 is set of of vertices which in G which are not isolated. Anyways those isolated vertices makes no contribution in strongly connected component.. so even if we choose those or not the graph gonna have the same connected component as that of G... Isn't it..??